Ask question

Ask question

All categories

3 years ago

11

The gravitational force between two objects has a magnitude of F. If both masses were doubled and the distance between them doub

led, the new force would have what magnitudes?

1 answer:

Fofino [41]3 years ago

7 0

Answer:

F' = F

Explanation:

The gravitational force of attraction between two objects can be given by Newton's Gravitational Law as follows:

$F = \frac{Gm_1m_2}{r^2}$

where,

F = Force of attraction

G = Universal gravitational costant

m₁ = mass of first object

m₂ = mass of second object

r = distance between objects

Now, if the masses and the distance between them is doubled:

$F' = \frac{G(2m_1)(2m_2)}{(2r)^2}\\\\F' = \frac{Gm_1m_2}{r^2}$

<u>F' = F</u>

You might be interested in

The instantaneous velocity of an object is the blank of the object with a blank

miss Akunina [59]

Instantaneous velocity is the velocity at a specific instant in time. I bet you are taking Honors Physics.

3 0

3 years ago

when an object is charged by contact, what kind of charge does the object have compared with the charge on the on the object giv

sveta [45]

Answer:

the charge that is given by the object is positive charge and the object which is taking the charge is negetively charged

Explanation:

5 0

3 years ago

In order for maximum work to be done...

Tju [1.3M]

free lil peezy and drake is kool

5 0

3 years ago

Read 2 more answers

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

<u>Now the horizontal component of velocity:</u>

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

<u>The vertical component of the velocity:</u>

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

<u>Time taken by the projectile to travel the distance of 30 m:</u>

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

<u>Vertical position of the projectile at this time:</u>

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

<u>We observe that </u> $t>t'$ <u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

8 0

3 years ago

A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to acc

Jlenok [28]

Answer:

a is the answer

Explanation:

3 0

3 years ago

Read 2 more answers